I wanted to start posting in probability but brotupelo has been a horrible poster to get me started. Constantly insulting everyone and just being absurd.

I reported his posts twice, and mods have edited or deleted them with no problems for brotupelo.

Anyone want to explain to me why he has no rules to follow and can act however he wants?

Because he is a free individual? I don't know. Is this a trick question?

But seriously, just ignore someone if they bother you. It's the internet and this is free public forum.

Is it a trick question? No. That is how forums work. You see useful debate and argue points.

Generally people are banned from forums(No, they aren't a free-speech democracy) when all they do is verbally abuse people and detract from discussion.

He made it literally impossible to try to have a civil debate about topics and points because he was constantly abusing people and "trolling" as he mentioned many times he was purposely doing.

This isn't BBV4L, it is supposed to be a forum where you can have useful discussions. I came and wanted to participate in some, but instead watched him degrade everyone. Just generally things that aren't allowed in any serious forum

In regards to the trick question, I was joking.

In regards to the rest, from what I can tell, it is just this one thread. I don't really think this is anything to get too upset over. If enough people don't like it and complain it will result in a ban.

There are hundreds of other threads in the probability forum you can read. If this one makes you unhappy, stop reading it. It really isn't that interesting of a probability question anyway (imo).

FWIW, the last 4 threads I've seen him post in he does the same thing. Go check out his posts in SMP.

Well, that is different. I don't read SMP (whatever that is), and I don't follow posters around to see what else they are posting. Again...if people complain enough he'll get banned. The squeaky wheel gets the grease. So if you don't like it, start squeaking. And if you still don't like it, squeak louder.

If we call A=(26,inf), and B=(0,inf), then we have:

B = A or (0,26]

If B is defined as A, plus some extra stuff, how can you possibly say they are the same size?

Because you can't add on to infinity

They are. There are several proofs that the number of reals between 0 and 1 is the same as the number of all reals, starting with Cantor's theorums. These two sets have the same cardinality. **

A good summary of infinite sets is here:
http://www.earlham.edu/~peters/writing/infapp.htm#thm13

Our intuition of how sets work breaks down when infinity is involved.

---
[I]** this makes some of the "solutions" offered ITT wrong

So the answer of the question in the first post is... ?

Sets are said to have the same cardinality (or informally "size") if and only if their elements can be placed into 1-to-1 correspondence with each other. In this case, each element a of set A can be paired with 1 unique element b of set B where b = a - 26, and each element b of set B can be paired with 1 unique element a of set A where a = b + 26. These sets have the same cardinality.

Finite sets have the same cardinality if and only if they have the same number of elements. For transfinite sets (sets with infinitely many elements) the cardinality of a set can be the same as the cardinality of a subset of that set, as in the example above. The sets of integers, even integers, odd integers, positive integers, negative integers, prime integers, and even rational numbers all have the same cardinality because it is possible to place their elements into 1-to-1 correspondence. These sets are said to be countably infinite because their elements can be put into 1-to-1 correspondence with the counting numbers (1,2,3,....), or "counted". It can be shown that the real numbers cannot be placed into 1-to-1 correspondence with the integers or the counting numbers, so the set of real numbers has a different cardinality, and is said to be uncountably infinite. Any interval of the real numbers has the same cardinality as the entire set of real numbers.

This kind of behavior will not be tolerated. Posts with insults or vulgarities will be and have been deleted. If this continues, there will be a single warning, followed by a banning if it still continues. Certainly a warning was warranted here. It wasn't issued yet, but it will be if I have to delete one more post.

finally...

BruceZ - are all infinite sets one of two sizes (cardinalities): countably infinite e.g. integers, and uncountably infinite e.g. reals? Or are there others? Do we differentiate countably infinite cardinalities according to proper subsets being smaller than the whole?

There are infinitely many uncountable infinities. The set of all subsets of any set (the power set) always has a higher cardinality than the set itself. This is proven by Cantor's diagonalization just as it is proven that the integers and reals have different cardinalities.

Now as to whether there exists a set with a cardinality between that of the integers and that of the reals, the continuum hypothesis states that there does not, and it is consistent to assume this hypothesis to be either true or false.

There is only 1 "countably infinite" cardinality. All infinite subsets of a countably infinite set are countably infinite.

spade: there are larger ones than that. In fact, given a set S, you can always construct a set of larger cardinality by considering the set of all maps f:S -> {0,1}. (which is called 2^S for natural reasons). You can show S sits inside 2^S by considering a map which takes x in S to a map f_x where f_x(x)=1 and f_x(y)=0 for all y<>x. Each map f_x is different, so 2^S is at least as big as S.
However, suppose you had a bijection (one-to-one correspondence) G between S and 2^S. Then you construct a map f:S -> {0,1} by letting f(x) = 0 if g(x)(x) = 1 and f(x) = 1 if g(x)(x) = 0. f is an element of 2^S, but it isn't in the image of G: if G(x) = f for some x, then we'd have G(x)(x) = f(x), which contradicts how we defined f. So, G can't exist, because no matter how we define it, we can find an element of 2^S that doesn't get hit. That means 2^S is strictly bigger than S. This works for any finite or infinite S, and is actually really similar to Cantor's proof that the reals are uncountable.

p.s. I don't know anything about cardinals bigger than 2^2^...^2^\aleph_0. Do you have to assume additional stuff to have them exist?

Alright this may just be a problem of notation, since I'm not a mathematician.

Is there another concept besides cardinality that relates to what I mean? Say we define A and B like this (over the integers, for simplicity's sake):

A = [26, N)
B = [0, N)

Now B contains N elements and A contains N-26 elements. So the difference in size between B and A is 26. If you take the limit as N goes to infinity then that difference should still be 26.

What I'm getting at is that if there are elements in B that aren't in A, but no elements in A that aren't in B, then in some sense B is "bigger" than A, even if they have the same cardinality. What's the term for that?

I don't know what I'd say other than A is a subset of B. It is enough to tell you that P(x in A) <= P(x in B), since P(x in [0, 26)) + P(x in [26, N)) = P(x in [0, N)) regardless of whether N is finite or infinity, and in that sense you can take the limit. The *reason* you can take the limit, though, is because the total probability is 1, so everything has to remain finite.

The difficulty of thinking about it in terms of size (or technically, Lebesgue measure is what I think you're getting at) is because you could say things like,
A = [26, N)
B = [0, 2N)
Now B is bigger than A, and B-A has measure N+26, but as N goes to infinity, should B-A have measure infinity, or measure 26?
But if we were taking the real line as a probability space, then P(B-A) does indeed converge to P([0, 26)) because everything is finite again.
I'm really not sure if this is answering your question though...

Ok yes that seems to be what I am thinking about, thanks.
If N goes to infinity, then N+26 should also go to infinity, no?


Not sure what you meant by this last bit...

If I understand the cardinality test correctly, the bolded part is the mistake. If N goes to infinity, for every number in B there will always be a number in A that can be matched up 1:1 with it. And vice versa. They are just offset. Start out pairing 26 with 0, pair 27 with 1, pair 28 with 2, and continue forever. They will always match, with none left over, and thus are the same size.

Now compare this with trying to match up the naturals with the reals. I can match up 1 with 1/1, 2 with 1/2, 3 with 1/3, 4 with 1/4 and keep going to infinity and I've covered all the naturals and matched them with a real. But there are still infinitely many more reals I'll never touch in this sequence. Thus the set of reals is larger than the set of naturals.

N+26 does go to infinity, but then what's your definition of the limiting set? To make this more precise, let's call them A_N = [26, N), B_N = [0, 2N). If m is Lebesgue measure, then
m(B_N) = 2N
m(A_N) = N-26
m(B_N - A_N) = N+26
It sounds like you want it to be ok to take the limit of (B_N-A_N), which you can from a set theoretic perspective: both lim_sup (B_N-A_N) = lim_inf (B_N-A_N) = [0,26). But then the measure of the limit is 26, and the limit of N+26 is infinity. So, this doesn't work, and the reason it doesn't work is because m(R) = infinity.
If we're looking at some probability measure on the reals though, then P(R) = 1, and you don't run into this problem because P(B_N - A_N) = P([0, 26)) + P([N, 2N)), and P([N, 2N)) has to go to 0. (If it didn't, we could find sets further and further out with positive probability that would eventually add up to more than 1.)

Just because one method fails to produce a 1-to-1 correspondence doesn't mean that the cardinalities are different. You must prove that a 1-to-1 correspondence is impossible.

It also sounds like you are confusing reals with rationals (p/q where p and q are integers) which CAN be counted similarly to what you are doing here. The reals cannot be counted because for any claimed numbered list of reals, I can specify a number which, for every N, differs from the Nth element of the list at the Nth decimal digit. This number cannot appear on the list, so no numbered list of reals can contain every real number.

No, it was just an easy example since all rationals are reals. My point was showing we can pair a unique real up with every natural and still have an infinite number of reals left over.

Thanks for the cardinality clarification.

Another thing that might clarify is Schroeder-Bernstein. It says, if I pair everything in A up with a unique thing in B, then #(A)<=#(B), regardless of how many things are left over in B from my pairing. If I take a map from R to R^2 which sends x to (x,0), this shows #(R) <= #(R^2), but doesn't tell me whether it's = or <. Once I pair everything in R^2 with a unique element of R by some other map, then I can conclude #(R^2)<=#(R), and so they must have the same cardinality.

By the Cantor-Bernstein-Schroeder theorem, if there are two injections ( 1-1 functions ) f: A → B and g: B → A, then A and B are equipotent or have the same cardinality. If R is the set of reals, A = R and B = RxR, there is the "obvious" injection from A to B; one choice for an injection in the other direction is to use a unique decimal representation ( don't use the one terminating in ...999... ) of reals and "interweave" the digits.

DarkMagus does raise an interesting point: how does a measure realize there is a "difference" between two subsets of R such as [0, infinity) and [1, infinity). The counting measure ( cardinality ) of each of these sets is the cardinality of the continuum and the Lebesgue measure ( "best" generalization of length on the real line ) of each of these sets is infinity. If the real numbers are made into a probability space, then the probability measure of the entire space is 1 and one can choose any probability measure that corresponds to a random variable X(.) for which on any nonempty interval I, P( X is in I ) > 0. Unfortunately, there isn't a "uniform" probability measure on all the reals, so the measure corresponding to using the standard normal is just about as good as any. In any case, by using such a probability measure, there will be a "difference" if one set ( that is measurable ) contains the other and the set difference contains some nonempty interval.